用單鏈表標示集合,設計一個算法求兩個結合的交集。
#include "stdafx.h"
#include "math.h"
typedef int ElemType;
typedef struct LNode
{
ElemType data;
struct LNode *next;
}LinkList;
//求LA和LB的交集LC
void interSet(LinkList *LA,LinkList *LB, LinkList *&LC)
{
//a的移動節點ta b的移動節點tb c的移動節點tc c的尾節點rc
LinkList *ta = LA->next, *tb,*tc,*rc;
//建立c頭
LC = new LinkList();
rc = LC;
while (ta)
{
tb = LB ->next;
while (tb && tb -> data != ta ->data)
tb = tb ->next;
if(tb)
{
tc = new LinkList();
tc ->data = ta ->data; //僅複製數據
rc->next = tc;//添加到c
rc = tc;//移動尾節點
}
ta = ta ->next;
}
rc ->next = 0;
}
void createLinkList(LinkList *&LA, ElemType data[], int n)
{
LA = new LinkList();
LinkList *ra = LA, *ta;
for (int idx= 0; idx < n; idx++)
{
ta = new LinkList();
ta ->data = data[idx];
ra -> next = ta;
ra = ta;
}
ra = 0;
}
int _tmain(int argc, _TCHAR* argv[])
{
LinkList *LA, *LB, *LC;
ElemType dataA[] = {5,4,6,1,8,3};
ElemType dataB[] = {8,5,9,12,4,7,6,3};
createLinkList(LA, dataA, 6);
createLinkList(LB, dataB, 8);
interSet(LA, LB, LC);
LinkList *tc = LC ->next;
while (tc)
{
printf("%4d", tc->data);
tc = tc -> next;
}
return 0;
}
運行結果:
ps:算法時間複雜度 O(n2), 想優化一下時間複雜度。想的結果是先讓a簡歷一個二叉樹,再比較B的元素。然後想爲何答案也是n2, 應該是沒有到二叉樹的知識吧。